This worksheet is a piece of cake. Hopefully this will get you started.

-qbkr21

May 9th 2007, 09:33 AM

alinailiescu

1 Attachment(s)

#1.(a)
f'(x)=3x^2-6x
f"(x)=6x-6
(b)see attachment

May 9th 2007, 09:43 AM

qbkr21

Re:

Quote:

Originally Posted by alinailiescu

#1.(a)
f'(x)=3x^2-6x
f"(x)=6x-6
(b)see attachment

Alinaliescu your differentiation and integration for part A & B are wrong

you were asked to integrate

f(x)=x^3+3x^2+5 NOT:eek: f(x)=x^3-3x^2+5

May 9th 2007, 10:00 AM

alinailiescu

Yes, I realized that after I posted the replay. I tried to print out the problems, and the printer didn't work properly.
Anyway, you posted the right solutions. :)
Thanks

May 9th 2007, 10:03 AM

qbkr21

Re:

I can see how you misread it. The vast majority of these PDF files people attach are very hard to read because of their clarity.

May 9th 2007, 10:10 AM

alinailiescu

#3
Since the points given are the end points of the diameter, the midpoint would be the center of the circle.
((-1+3)/2, (4+6)/2)
thus, (1, 5) is the center.
You also need the radius,
r=sqrt[(3-1)^2+(6-5)^2]=sqrt(5)
So, the equation is:
(x-1)^2+(y-5)^2=5

May 9th 2007, 10:25 AM

novadragon849

Thanks so is anyone able to do questions 6-10

May 9th 2007, 10:36 AM

alinailiescu

#6

first ,use cos^2(x)=1-sin^2(x) to eliminate cosine function
That would give you:
2(1-sin^2(x))+1=5sinx
2-2sin^2(x)+1-5sinx=0
-2sin^2(x)-5sinx+3=0
This is a quadratic equation in sinx, so using the quadratic formula you have:
sinx=[5+sqrt(25-4*-2*3)]/[-2*2]=[5+7]/(-4)=12/-4=-3(which has no solution for x, since sinx should be between -1 and 1)
or
sinx==[5-sqrt(25-4*-2*3)]/[-2*2]=[5-7]/(-4)=1/2
so x=sin^-1(1/2)
x=pi/6